MATH 203 one-sided limits: revision sheet

Q: How do I find a left-hand limit with an absolute value?

Look at the inside of the absolute value just to the left of the point. If it is negative there, replace the absolute value by minus the inside; if it is positive, by the inside itself. Then simplify and take the limit of that formula. Do the same on the right with the sign that holds there, and compare the two answers.

Q: When does a limit not exist in calculus 1?

In this chapter, a two-sided limit fails to exist when the left-hand and right-hand limits are two different numbers, or when one of them does not exist. It does not fail just because the function is undefined at the point: a hole still has a limit. At the endpoint of a domain only one side can be computed, and that one-sided limit is the answer.

Q: Why does sin x over x tend to 1 but not when x goes to pi?

The theorem proved in Thomas section 2.4 is about an angle that tends to zero, measured in radians. When x tends to pi, the sine tends to zero but the denominator tends to pi, so there is no indeterminate form and the limit is simply zero over pi, which is zero. Always substitute first: the theorem is only needed for zero over zero.

Q: What is the limit of the floor function at an integer?

At an integer n, the floor of x equals n minus one just to the left of n and n just to the right, so the left-hand limit is n minus one and the right-hand limit is n. Both one-sided limits exist, but they are different, so the two-sided limit does not exist. Between two integers the floor is constant and its limit is that constant.

The thread of the chapter

Where the formula changes (the inside of $|u|$ vanishes, a floor meets an integer, a junction, an endpoint), the two sides are two different functions: each side is computed with ITS formula, and the marks are lost in the algebra that picks that formula.

Topics covered

One-sided limits
Left-hand limit
Right-hand limit
Limits at domain endpoints
Absolute value function
Greatest integer function
Least integer function
Piecewise-defined function
Fundamental trigonometric limit
Half-angle identity
Double-angle identity

The essentials

Two sides, two formulas

• $\lim_{x\to a} f(x) = L$ if and only if $\lim_{x\to a^-} f(x) = L$ AND $\lim_{x\to a^+} f(x) = L$ . The value $f(a)$ plays no part.
• Split into two sides only where the formula CHANGES at $a$ : the inside of $|u|$ vanishes at $a$ , the argument of $\lfloor u \rfloor$ or $\lceil u \rceil$ is an integer at $a$ , a piecewise function has a junction at $a$ .
• $|u| = u$ where $u \ge 0$ , $|u| = -u$ where $u < 0$ ; and $\sqrt{u^2} = |u|$ , never $u$ .
• At an endpoint of the domain only the side inside the domain exists: $\lim_{x\to 0^+} \sqrt x = 0$ , and there is no two-sided limit there (Thomas 2.4).
• $\lfloor u \rfloor$ is constant on $[n, n + 1)$ , $\lceil u \rceil$ on $(n, n + 1]$ : read the interval of the ARGUMENT $u$ on each side.

For

\frac{x^2 - x}{|x - 1|}

the left side follows

y = -x

towards

-1

and the right side follows

y = x

towards

1

: two formulas, two targets, no limit at

1

Write the sign of the inside on each side as a sentence (“for $x < 3$ , $x - 3 < 0$ , so $|x - 3| = -(x - 3)$ ”): it is the line the marker is looking for.

The two trigonometric limits of Thomas 2.4, and the identities that feed them

• $\lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1$ and $\lim_{h\to 0} \frac{\cos h - 1}{h} = 0$ , angles in RADIANS, the angle tending to $0$ .
• The angle may be anything that tends to $0$ : $3x$ , $\sqrt x$ , $x - 1$ , $x^2 - 5x$ . The denominator must be made EQUAL to it, by multiplying and dividing by the same constant or by factoring.
• Reciprocal: $\frac{\theta}{\sin \theta} \to 1$ too, by the quotient rule, since $1 \ne 0$ .
• Identities: $\tan u = \frac{\sin u}{\cos u}$ , $\sin 2u = 2\sin u \cos u$ , $1 - \cos 2u = 2\sin^2 u$ , $\sin^2 u = (1 - \cos u)(1 + \cos u)$ .
• If substitution gives no $\frac{0}{0}$ , there is nothing to rewrite: $\lim_{x\to \pi} \frac{\sin x}{x} = \frac{0}{\pi} = 0$ .

y = \frac{\sin x}{x}

x

in radians, is undefined at

0

and yet climbs to

1

from both sides:

\lim_{x\to 0} \frac{\sin x}{x} = 1

A constant never leaves a trigonometric function or a floor: $\tan 3x \ne 3\tan x$ , $\sin 2x \ne 2\sin x$ , $\lfloor 2x \rfloor \ne 2\lfloor x \rfloor$ .

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Opening an absolute value with the wrong sign on the left

the whole question

What not to write

“ $\frac{x^2 - 9}{|x - 3|} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3$ , so the limit at $3$ is $6$ .”

What to write

“For $x > 3$ , $|x - 3| = x - 3$ and the quotient is $x + 3 \to 6$ . For $x < 3$ , $|x - 3| = -(x - 3)$ and the quotient is $-(x + 3) \to -6$ . The sides differ: the limit does not exist.”

\frac{x^2 - 9}{|x - 3|}

x + 3

on the right (towards

6

) and

-(x + 3)

on the left (towards

-6

): forgetting the minus sign on the left makes the two sides agree by accident.

Why: Where the inside vanishes at $a$ , it is negative on one side. An absolute value is never “just removed”: it is opened with the sign of its inside, side by side.

2. Splitting where the inside of the absolute value does not vanish

the whole question

What not to write

“At $-3$ , $|x| = x$ on the right and $-x$ on the left, so $\frac{|x| - 3}{x + 3}$ has two different one-sided limits.”

What to write

“Near $-3$ , $x < 0$ on both sides, so $|x| = -x$ and $\frac{|x| - 3}{x + 3} = \frac{-x - 3}{x + 3} = -1$ : the limit is $-1$ .”

Around

-3

x < 0

on both sides: the graph of

\frac{|x| - 3}{x + 3}

is flat at

-1

, with a hole. Its only bend is at

0

, where

|x|

changes formula.

Why: The split happens where the INSIDE changes sign, here at $0$ , not where the limit is taken. Near $-3$ the sign of $x$ is fixed.

3. Taking the square root of a square without its absolute value

the whole question, often 3 marks

What not to write

“ $1 - \cos x = 2\sin^2\frac{x}{2}$ , so $\frac{\sqrt{1 - \cos x}}{x} = \frac{\sqrt 2 \sin(x/2)}{x} \to \frac{\sqrt 2}{2}$ .”

What to write

“ $\sqrt{2\sin^2\frac{x}{2}} = \sqrt 2 \left|\sin\frac{x}{2}\right|$ . For $x > 0$ the limit is $\frac{\sqrt 2}{2}$ ; for $x < 0$ , $|\sin\frac{x}{2}| = -\sin\frac{x}{2}$ and the limit is $-\frac{\sqrt 2}{2}$ . No two-sided limit.”

y = \frac{\sqrt{1 - \cos x}}{x}

approaches

\frac{\sqrt 2}{2}

from the right and

-\frac{\sqrt 2}{2}

from the left: the root hides an absolute value.

Why: A square root is never negative, while $\sin\frac{x}{2} < 0$ for small $x < 0$ . The absolute value hidden in $\sqrt{u^2}$ is what splits the sides.

4. Reading the floor of 2x on x

the whole question

What not to write

“ $\lfloor 2x \rfloor = 2\lfloor x \rfloor$ , and $\lfloor x \rfloor = -1$ near $-\frac{1}{2}$ , so $\lim_{x\to -1/2} \lfloor 2x \rfloor = -2$ .”

What to write

“The argument is $2x$ . For $x \to -\frac{1}{2}^-$ , $2x \in [-2, -1)$ and $\lfloor 2x \rfloor = -2$ ; for $x \to -\frac{1}{2}^+$ , $2x \in [-1, 0)$ and $\lfloor 2x \rfloor = -1$ . No limit.”

\lfloor 2x \rfloor

(blue steps) jumps at

-\frac{1}{2}

and

\frac{1}{2}

, where the dashed

2\lfloor x \rfloor

stays flat: two different functions.

Why: $\lfloor 2x \rfloor$ jumps at every half-integer, $2\lfloor x \rfloor$ only at integers. A floor is read on its own argument, as $\lfloor x^2 \rfloor$ is read on $x^2$ .

5. Reading a one-sided limit on the value at the jump

2 marks

What not to write

“The parking price is $P(3) = 10$ dollars, so $\lim_{t\to 3^+} P(t) = 10$ .”

What to write

“For $3 < t \le 4$ , $P(t) = 13$ , so $\lim_{t\to 3^+} P(t) = 13$ ; the value $P(3) = 10$ sits on the left step.”

P(3) = 10

sits on the left step (full dot); just after

3

the price is

13

(empty dot). The limit from the right reads the step on the right, not the dot.

Why: A right-hand limit uses only $t > 3$ . With a ceiling the value at the integer belongs to the LEFT step, with a floor to the right one: read the interval, not the dot.

6. Matching the sine with the wrong denominator

2 marks

What not to write

“ $\frac{\sin(3x/4)}{5x} \to \frac{1}{5}$ , since the sine over $x$ tends to $1$ .”

What to write

“ $\frac{\sin(3x/4)}{5x} = \frac{3}{20} \cdot \frac{\sin(3x/4)}{3x/4}$ and $\frac{3x}{4} \to 0$ , so the limit is $\frac{3}{20}$ .”

Why: The theorem needs the SAME quantity in the sine and in the denominator. Dividing by $\frac{3x}{4}$ is multiplying by $\frac{4}{3x}$ : the compound fraction is where the constant goes wrong.

7. Quoting sin(θ)/θ → 1 when the angle does not tend to 0

the whole question

What not to write

“ $\lim_{x\to \pi} \frac{\sin x}{x} = 1$ , it is sine over its angle.”

What to write

“Substitution gives $\frac{\sin \pi}{\pi} = \frac{0}{\pi}$ , no indeterminate form: by the quotient rule the limit is $0$ .”

x = \pi

the curve

y = \frac{\sin x}{x}

is at height

0

(green), not

1

(red): substitution gives

\frac{0}{\pi}

, no indeterminate form, nothing to rewrite.

Why: The theorem is about an angle tending to $0$ . Substitute first: without a $\frac{0}{0}$ there is nothing to rewrite.

8. Treating the cosine limit like the sine limit

2 marks, although the rewriting is right

What not to write

“ $\frac{\cos 5h - 1}{\sin 2h} = \frac{\cos 5h - 1}{5h} \cdot \frac{5h}{2h} \cdot \frac{2h}{\sin 2h} \to 1 \cdot \frac{5}{2} \cdot 1 = \frac{5}{2}$ .”

What to write

“The same rewriting, but $\frac{\cos 5h - 1}{5h} \to 0$ : the limit is $0 \cdot \frac{5}{2} \cdot 1 = 0$ .”

Why: Thomas 2.4 proves $\frac{\cos h - 1}{h} \to 0$ , not $1$ : it follows from $\cos h - 1 = -2\sin^2\frac{h}{2}$ , a product with a factor tending to $0$ .

9. Taking a limit on a side where the function has no values

1 mark

What not to write

“ $\lim_{x\to 0^-} \sqrt x = 0$ and $\lim_{x\to 0^+} \sqrt x = 0$ , so $\lim_{x\to 0} \sqrt x = 0$ .”

What to write

“ $\sqrt x$ is defined only for $x \ge 0$ : $\lim_{x\to 0^+} \sqrt x = 0$ , there is no left-hand limit, and the two-sided limit does not exist at this endpoint.”

\sqrt x

has no points left of

0

(grey band): only

\lim_{x\to 0^+} \sqrt x = 0

exists, and there is no two-sided limit at this endpoint.

Why: A one-sided limit needs values on that side. At an endpoint, Thomas asks the one side inside the domain, and that side is the full answer.

Which method to choose

Which gesture, by the FORM of the limit

Substitute a first, on the side. What you see picks the tool

If $|u|$ with $u(a) = 0$ → split: $|u| = u$ on the side where $u > 0$ , $-u$ on the other; compute both sides

Example: $\frac{2x^2 - 5x}{|2x - 5|}$ at $\frac{5}{2}$ : $-\frac{5}{2}$ and $\frac{5}{2}$ , no limit
If $|u|$ with $u(a) \ne 0$ → no split: the sign of u is the same on both sides, open the absolute value once

Example: $\frac{|x| - 3}{x + 3} = -1$ near $-3$
If $\sqrt{u^2}$ , or a root of $1 - \cos$ → write $|u|$ first, then treat it as an absolute value

Example: $\frac{\sqrt{1 - \cos 2x}}{x}$ : $\sqrt 2$ and $-\sqrt 2$
If $\lfloor u \rfloor$ or $\lceil u \rceil$ with $u(a)$ an integer → split; on each side replace the floor by the integer of the interval of u

Example: $\lfloor x \rfloor^2 - \lfloor x^2 \rfloor$ at $3$ : $-4$ and $0$
If a junction of a piecewise function → each side with the formula valid on that side; the value at the junction is not used

Example: $x^2 + b$ on the left of $2$ , $x + 1$ on the right: $b = -1$
If $\sin(\text{angle})$ over something, the angle tending to $0$ → multiply and divide so that the denominator equals the angle; factor it if needed

Example: $\frac{\sin(x^2 - 5x)}{2x} \to -\frac{5}{2}$
If $1 - \cos$ or $\cos - 1$ in a $\frac{0}{0}$ → half angle $1 - \cos 2u = 2\sin^2 u$ , or Pythagoras $\sin^2 = (1 - \cos)(1 + \cos)$

Example: $\frac{1 - \cos 6x}{x \sin 3x} \to 6$
If no 0/0 after substitution → done: quotient rule, no rewriting

Example: $\frac{\sin x}{x} \to 0$ at $\pi$

Infinite limits and limits at infinity are the next chapter; if a side tends to a nonzero number over $0$ , stop at “no finite limit” for now.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Proving that a limit does not exist with one-sided limits

When to use it: An absolute value whose inside vanishes at $a$ , a floor at an integer, a junction

1 Say why the sides must be separated: “the inside $x - 3$ vanishes at $3$ ”, or “the formula changes at $2$ ”.
2 Left side: state the sign or the interval (“for $x < 3$ , $x - 3 < 0$ ”), write the formula valid there, take its limit.
3 Right side: the same, with its own formula.
4 Compare the two numbers and conclude with the reason.

Concluding sentence

“For $x < 3$ , $|x - 3| = -(x - 3)$ , so $\lim_{x\to 3^-} \frac{x^2 - 9}{|x - 3|} = -6$ ; for $x > 3$ , $|x - 3| = x - 3$ , so $\lim_{x\to 3^+} \frac{x^2 - 9}{|x - 3|} = 6$ . The one-sided limits are different, so the limit does not exist.”

The trap: Writing “the limit does not exist because $f(3)$ is undefined”: that is never a reason, a hole has a limit.

Marking: Typically 1 mark per one-sided limit with its sign sentence, 1 for the conclusion with its reason.

Rewriting a limit into sin(θ)/θ, written for full marks

When to use it: A sine (or a tangent) of an angle tending to $0$ , in a form $\frac{0}{0}$

1 Substitute and write the form $\frac{0}{0}$ .
2 Name the angle $\theta$ and check that $\theta \to 0$ .
3 Multiply and divide so that $\frac{\sin \theta}{\theta}$ appears, with every constant written.
4 Quote the theorem, then the product or quotient rule, and conclude.

Concluding sentence

“With $\theta = x - 1 \to 0$ : $\frac{\sin(x - 1)}{x^2 - 1} = \frac{\sin(x - 1)}{x - 1} \cdot \frac{1}{x + 1} \to 1 \cdot \frac{1}{2} = \frac{1}{2}$ , by $\lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1$ and the product rule.”

The trap: Cancelling the sine, as in $\frac{\sin 3x}{\sin 8x} = \frac{3x}{8x}$ : the number may survive, the method mark does not.

Marking: Usually 1 mark for the rewriting, 1 for the theorem named, 1 for the value.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

A table on each side, in radian mode

The department's scientific calculator computes no limit, but it evaluates. Take x = a - 0.001 and x = a + 0.001, in RADIAN mode, and compare with your two answers.

$\frac{\sqrt{1 - \cos 2x}}{x}$ at $x = \pm 0.001$ gives about $\pm 1.4142$ : two answers, as the absolute value predicts. In degree mode you get nonsense.
The sign on one side

Take x just to the left of a, read the sign of every factor. A left-hand limit cannot have the opposite sign.

$\frac{2x^2 - 5x}{|2x - 5|}$ at $x = 2.4$ : $\frac{-0.48}{0.2} = -2.4$ , so the left limit is negative, near $-2.5$ .
Substitute before quoting a theorem

If the substitution gives a number over a nonzero number, the answer is that quotient. The trigonometric limits are for 0/0 only.

$\frac{\sin x}{x}$ at $\pi$ : $\frac{0}{\pi} = 0$ , not $1$ .
Test an identity at one angle

Before using an identity you are not sure of, try it at an easy angle.

$1 - \cos 2u = 2\sin^2 u$ at $u = \frac{\pi}{2}$ : $1 - (-1) = 2$ and $2 \cdot 1 = 2$ . The unhalved version $2\sin^2 2u$ gives $0$ : wrong.

The typical problem, taken apart

A junction with a sine on one side and an absolute value on the other

Let $f(x) = \frac{\sin(kx)}{4x}$ for $x < 0$ , $f(0) = 5$ , and $f(x) = \frac{x^2 + 3x}{2|x|}$ for $x > 0$ , where $k$ is a constant.

Find $k$ so that $\lim_{x\to 0} f(x)$ exists, give the limit, and compare it with $f(0)$ .

Step 1

The formula changes at $0$ : compute $\lim_{x\to 0^-} f(x)$ and $\lim_{x\to 0^+} f(x)$ separately. The value $f(0) = 5$ is set aside.

Why

A junction forces the split. Using $f(0)$ now would answer a different question.

Step 2

Right: for $x > 0$ , $|x| = x$ , so $f(x) = \frac{x(x + 3)}{2x} = \frac{x + 3}{2}$ and $\lim_{x\to 0^+} f(x) = \frac{3}{2}$ .

Why

Saying $|x| = x$ BECAUSE $x > 0$ is the justification; factoring $x^2 + 3x = x(x + 3)$ is the algebra that makes the cancellation visible.

Step 3

Left: for $k \ne 0$ , $\frac{\sin(kx)}{4x} = \frac{k}{4} \cdot \frac{\sin(kx)}{kx}$ and $kx \to 0$ , so $\lim_{x\to 0^-} f(x) = \frac{k}{4}$ (and $0 = \frac{k}{4}$ if $k = 0$ ).

Why

The angle is $kx$ , so the denominator is made into $kx$ ; the constant $\frac{k}{4}$ is written, not guessed.

Step 4

The limit exists if and only if $\frac{k}{4} = \frac{3}{2}$ , that is $k = 6$ . Then $\lim_{x\to 0} f(x) = \frac{3}{2}$ , while $f(0) = 5$ .

Why

The constant makes the SIDES agree; the value at $0$ plays no part, and here it differs from the limit.

Step 5

Check with the calculator in radian mode: $x = -0.01$ gives $\frac{\sin(-0.06)}{-0.04} \approx 1.4991$ ; $x = 0.01$ gives $\frac{3.01}{2} = 1.505$ . Both are near $1.5$ .

Why

A value on each side catches a lost constant or a wrong sign in the absolute value at no cost.

The conclusion, written out

“With $k = 6$ , $\lim_{x\to 0^-} f(x) = \lim_{x\to 0^+} f(x) = \frac{3}{2}$ , so $\lim_{x\to 0} f(x) = \frac{3}{2}$ , although $f(0) = 5$ .”

The classic mistake on this problem: Choosing $k = 4$ so that $\frac{\sin(kx)}{4x}$ “looks like” $\frac{\sin \theta}{\theta}$ : the left limit is then $1$ , not $\frac{3}{2}$ .

Learn by heart

• $\lim_{x\to a} f(x) = L$ iff both one-sided limits exist and EQUAL $L$ . The value $f(a)$ plays no part.
• Split only where the formula changes: inside of $|u|$ zero at $a$ , integer argument of a floor or ceiling, junction.
• $|u| = -u$ where $u < 0$ . $\sqrt{u^2} = |u|$ .
• At an endpoint, only the side inside the domain exists: $\lim_{x\to 0^+} \sqrt x = 0$ .
• $\frac{\sin \theta}{\theta} \to 1$ and $\frac{\cos h - 1}{h} \to 0$ , radians, angle $\to 0$ , same angle above and below.
• $1 - \cos 2u = 2\sin^2 u$ , $\sin 2u = 2\sin u \cos u$ , $\sin^2 u = (1 - \cos u)(1 + \cos u)$ .
• Constants never leave a function: $\lfloor 2x \rfloor \ne 2\lfloor x \rfloor$ , $\tan 3x \ne 3\tan x$ .

Frequently asked questions

How do I find a left-hand limit with an absolute value?

Look at the inside of the absolute value just to the left of the point. If it is negative there, replace the absolute value by minus the inside; if it is positive, by the inside itself. Then simplify and take the limit of that formula. Do the same on the right with the sign that holds there, and compare the two answers.

When does a limit not exist in calculus 1?

In this chapter, a two-sided limit fails to exist when the left-hand and right-hand limits are two different numbers, or when one of them does not exist. It does not fail just because the function is undefined at the point: a hole still has a limit. At the endpoint of a domain only one side can be computed, and that one-sided limit is the answer.

Why does sin x over x tend to 1 but not when x goes to pi?

The theorem proved in Thomas section 2.4 is about an angle that tends to zero, measured in radians. When x tends to pi, the sine tends to zero but the denominator tends to pi, so there is no indeterminate form and the limit is simply zero over pi, which is zero. Always substitute first: the theorem is only needed for zero over zero.

What is the limit of the floor function at an integer?

At an integer n, the floor of x equals n minus one just to the left of n and n just to the right, so the left-hand limit is n minus one and the right-hand limit is n. Both one-sided limits exist, but they are different, so the two-sided limit does not exist. Between two integers the floor is constant and its limit is that constant.

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Practise it

Corrected exercises: One-sided limits, MATH 203 at Concordia

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

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Revision sheet: one-sided limits (MATH 203)

The essentials

Two sides, two formulas

The two trigonometric limits of Thomas 2.4, and the identities that feed them

The mistakes that cost marks

1. Opening an absolute value with the wrong sign on the left

2. Splitting where the inside of the absolute value does not vanish

3. Taking the square root of a square without its absolute value

4. Reading the floor of 2x on x

5. Reading a one-sided limit on the value at the jump

6. Matching the sine with the wrong denominator

7. Quoting sin(θ)/θ → 1 when the angle does not tend to 0

8. Treating the cosine limit like the sine limit

9. Taking a limit on a side where the function has no values

Which method to choose

Which gesture, by the FORM of the limit

How the answer is expected to be written

Proving that a limit does not exist with one-sided limits

Rewriting a limit into sin(θ)/θ, written for full marks

Check before you hand in

The typical problem, taken apart

A junction with a sine on one side and an absolute value on the other

Learn by heart

Frequently asked questions

How do I find a left-hand limit with an absolute value?

When does a limit not exist in calculus 1?

Why does sin x over x tend to 1 but not when x goes to pi?

What is the limit of the floor function at an integer?

Corrected exercises: One-sided limits, MATH 203 at Concordia

See also

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